■トリボナッチ数列(その25)

3次方程式の解を-1、ω、ω^-1とする。t0=1,t1=135,t2=11161,t3=926271

ω+ω^-1=83

ω・ω^-1=1

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α=-1

β=ω

γ=ω^-1

α+β+γ=-1+ω+ω^-1=82

αβ+βγ+γα=1-ω+ω^-1=-82

αβγ=-1

Un+1=Tn+1-αTn

U0=T1-αT0=136

U1=T2-αT1=11296

U2=T3-αT2=937432

Vn=Un+1-βUn

V0=U1-βU0=11296-136ω

V1=U2-βU1=937432-ω(11296-136ω)

Vn+1=γVn

Vn=ω^-n・(11296-136ω)

Wn=Un+1-γUn

W0=U1-γU0=11296-136ω^-1

W1=U2-γU1=937432-ω^-1(11296-136ω^-1)

Wn+1=βWn

Wn=ω^n・(11296-136ω^-1)

Un+1-βUn=ω^-n・(11296-136ω)

Un+1-γUn=ω^n・(11296-136ω^-1)

Un=(ω^n・(11296-136ω^-1)-ω^-n・(11296-136ω))/(ω-ω^-1)

Tn+1-αTn=(ω^n・(11296-136ω^-1)-ω^-n・(11296-136ω))/(ω-ω^-1)

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t0=1,t1=135,t2=11161,t3=926271

Un+1=Tn+1-βTn

U0=T1-βT0=135-ω

U1=T2-βT1=11161-135ω

U2=T3-βT2=926271-11161ω

Vn=Un+1-γUn

V0=U1-γU0=11161-135ω-ω^-1・(135-ω)=-43

V1=U2-γU1

Vn+1=αVn

Vn=(-1)^n・(-43)

Wn=Un+1-αUn

W0=U1-αU0=11161-135ω+(135-ω)=11296-136ω

W1=U2-αU1=

Wn+1=γWn

Wn=ω^-n・(11296-136ω)

Un+1-γUn=(-1)^n・(-43)

Un+1-αUn=ω^-n・(11296-136ω)

Un=(ω^-n・(11296-136ω)-(-1)^n(-43))/(ω^-1+1)

Tn+1-βTn=(ω^-n・(11296-136ω)-(-1)^n(-43)))/(ω^-1+1)

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t0=1,t1=135,t2=11161,t3=926271

Un+1=Tn+1-γTn

U0=T1-γT0=135-ω^-1

U1=T2-γT1=11161-135ω^-1

U2=T3-γT2=926271-11161ω^-1

Vn=Un+1-βUn

V0=U1-βU0=11161-135ω^-1-ω・(135-ω^-1)=-43

V1=U2-βγU1

Vn+1=αVn

Vn=(-1)^n・(-43)

Wn=Un+1-αUn

W0=U1-αU0=11161-135ω^-1+(135-ω^-1)=11296-136ω^-1

W1=U2-αU1=

Wn+1=βWn

Wn=ω^n・(11296-136ω^-1)

Un+1-βUn=(-1)^n・(-43)

Un+1-αUn=ω^n・(11296-136ω^-1)

Un=(ω^n・(11296-136ω^-1)-(-1)^n(-43))/(ω+1)

Tn+1-γTn=(ω^n・(11296-136ω^-1)-(-1)^n(-43)))/(ω+1)

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Tn=-{(β-γ)α^n+(γ-α)β^n+(α-β)γ^n}/(α-β)(β-γ)(γ-α)

α^n→(-1)^n・(-43)

β^n→ω^n・(11296-136ω^-1)

γ^n→ω^-n・(11296-136ω)に置換

(α-β)=-1-ω

(β-γ)=(ω-ω^-1)

(γ-α)=(ω^-1+1)

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n=0のとき

T0=-{(β-γ)(-43)+(γ-α)(11296-136ω^-1)+(α-β)(11296-136ω)}/(α-β)(β-γ)(γ-α)=1になるだろうか?

もっと簡単になればよいのであるが・・・x^2-83x+1=0

x=1/2・{83+-{83^2-4}^1/2}

x=1/2・{83+-6885^1/2}

x=1/2・{83+-82.9759・・・}

(ω-ω^-1)=(6885)^1/2

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